Initial time layer problem for quantum drift-diffusion model

J. Math. Anal. Appl. 343 (2008) 64–80 www.elsevier.com/locate/jmaa

Initial time layer problem for quantum drift-diffusion model ✩ Xiuqing Chen a,∗ , Li Chen b a School of Sciences, Beijing University of Posts and Telecommunications, Beijing 100876, China b Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China

Received 16 September 2007 Available online 15 January 2008 Submitted by P. Broadbridge

Abstract We study the initial time layer problem to the quantum drift-diffusion model. The limit of vanishing Debye length is established by entropy method and compactness argument. As a byproduct, the global weak solution is also obtained for this model. © 2008 Elsevier Inc. All rights reserved. Keywords: Initial time layer; Quantum drift-diffusion model; Compactness argument

1. Introduction The quantum macroscopic models, namely quantum drift-diffusion model (QDDM), quantum hydrodynamic model and quantum energy transport model were introduced recently to simulate the quantum effects in miniaturized semiconductor devices. Some derivation of these models could be found in [10,11,26,29] and some theoretical analysis for the latter two models was given in [4,5,12] etc. We will consider the bipolar QDDM, ⎧    √  α

n ⎪ 2 n∇  ⎪ ⎪ n + ∇ n − n∇V , = div −ε √ t ⎪ ⎪ n ⎪ ⎨    √  (1.1)

2 p∇  p + ∇ p β + p∇V , ⎪ p = div −ξ ε √ ⎪ t ⎪ p ⎪ ⎪ ⎪ ⎩ 2 λ V = n − p − C(x), where the electron density n, hole density p and the electrostatic potential V are unknown variables; the doping profile C(x) representing the distribution of charged background ions is supposed to be independent of time t; ξ > 0 is the ratio of the effective masses of electrons and holes; α, β  1, the scaled Planck constant ε > 0 and Debye length λ > 0 are parameters. ✩

Project supported by the National Natural Science Foundation of China (No. 10401019) and the Basic Research Grant of Tsinghua University.

* Corresponding author.

E-mail address: [email protected] (X. Chen). 0022-247X/$ – see front matter © 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2008.01.015

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

65

The main difficulty of such equations as (1.1) lies in the nonlinear fourth order parabolic equation and its feature of lack of maximum principle. Recently, a series of works [1–3,6–9,13–25] study the well-posedness of QDDM and its special case with zero electrostatic potential as well as the semiclassical or quasi-neutral limit in QDDM. If we introduce the fast time scale s = λt2 and Vλ = λ2 V , the system (1.1) turns into    √  ⎧

nλ ⎪ 2 λ2 n ∇  2 ∇ nα − n ∇V , ⎪ (n + λ ) = div −ε √ λ λ λ ⎪ λ s λ ⎪ ⎪ nλ ⎨    √  β

 pλ 2 2 2 ⎪ (pλ )s = div −ξ ε λ pλ ∇ √ + λ ∇ pλ + pλ ∇Vλ , ⎪ ⎪ ⎪ pλ ⎪ ⎩ Vλ = nλ − pλ − C(x),

(1.2)

where nλ , pλ and Vλ represent the new variables dependent on λ. In the literature, the limit as the Debye length λ → 0 is interesting, which is one of the initial time layer problems. Some authors performed this kind of limit for the drift-diffusion model, a system without fourth order quantum term, in [28–33] etc. But for the QDDM, a fourth order parabolic system, no one has checked this problem up to the authors’ knowledge. The main task of this paper is to investigate the limit for (1.2) with periodic boundary condition in two or three space γ −1 γ −1 dimensions (d = 2, 3). First, by using nγ −1 , pγ −1 (γ > 1) as test functions, we could establish new entropy inequality and hence obtain the global weak solution with more regularity than [25] when the exponents to the pressure are enhanced greatly to 1  α, β < 92 (d = 2) or 1  α, β < 16 5 (d = 3). In contrast, those exponents in [25] are limited to 1  α, β < 3 (d = 2) or 1  α, β  2 (d = 3). Then we should cope with the main difficulty of implementing the limit, that is how to build the compulsory uniform estimates especially those of products λγ1 f1 (nλ ), λγ2 f2 (pλ ). By employing delicate interpolation skill and chain rule, we solve this problem successfully based on the entropy inequalities. Since all the results in this paper are obtained for fixed ξ, ε > 0, for convenience we let ξ = ε = 1 in the following. To search for solutions which are physically reasonable, namely the solutions in which variables nλ and pλ are nonnegative, we might as well suppose nλ = ρλ2 and pλ = ηλ2 . Moreover, let S > 0 be any fixed constant and QS = (0, S] × Td where Td represents d-dimensional flat torus. We will consider the initial periodic boundary value problem in QS   ⎧   2

ρλ ⎪ 2α 2 2 2 2 ⎪ ρλ s = div −λ ρλ ∇ + λ ∇ρλ − ρλ ∇Vλ ⎪ ⎪ ⎪ ρλ ⎪ ⎪ ⎪ ⎪ d ⎪



⎪ λ2 2 ⎪ ⎪ =− ρλ ln ρλ2 x x x x + λ2 ρλ2α − div ρλ2 ∇Vλ , ⎪ ⎪ i j i j 2 ⎪ ⎪ i,j =1 ⎪ ⎪ ⎪     ⎪ ⎪ 2

⎪ ηλ 2β ⎪ 2 2 2 2 ⎪ ⎪ ⎪ ηλ s = div −λ ηλ ∇ ηλ + λ ∇ηλ + ηλ ∇Vλ ⎪ ⎨ d

λ2 2 2

2β ⎪ ⎪ = − ηλ ln ηλ x x x x + λ2 ηλ + div ηλ2 ∇Vλ , ⎪ ⎪ i j i j 2 ⎪ ⎪ i,j =1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Vλ = ρλ2 − ηλ2 − C(x), ⎪ ⎪ ⎪ ⎪ ⎪ ρλ (0, x) = ρ0 (x), ηλ (0, x) = η0 (x), ⎪ ⎪ ⎪

⎪ ⎪ 2

⎪ ⎪ ⎪ ρ0 − η02 dx = C(x) dx, ⎪ ⎪ ⎩ Td

where



2 Td (ρ0

(1.3)

Td

− η02 ) dx =

 Td

C(x) dx is a necessary condition such that the Poisson equation in (1.3) is solvable.

66

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

Notation. • The Sobolev spaces, W m,p (Td ) (H m (Td ) = W m,2 (Td )); the Hölder spaces, C k,θ (Td ); • H −2 (Td ) denotes the dual space of H 2 (Td ) and ·,· represents the duality pairing between H −2 (Td ) and H 2 (Td ); D 2 u represents the Hessian matrix of u; A → B (or A → → B) denotes A is continuously (or compactly) embedded in B; • γ − (or γ + ) means that γ − < γ (or γ + > γ ) and γ − (or γ + ) is close to γ sufficiently. Particularly, ∞− denotes the large enough real number. Our main results are stated as follows: ∈ L∞ (Td ). Suppose that ρ0 , η0 Theorem 1.1. Let d = 2, 1  α, β < 92 or d = 3, 1  α, β < 16 5 and C(x)   2 2 d are nonnegative measurable functions on T with Td (ρ0 − η0 ) dx = Td C(x) dx and ρ02 (ln ρ02 − 1) + 1, η02 (ln η02 − 1) + 1 ∈ L1 (Td ). Then for any fixed λ > 0, there exists a weak solution (ρλ , ηλ , Vλ ) to (1.3) in the following sense: Case i: d = 2, 1  α, β < 92 .



− 0  ρλ , ηλ ∈ L∞ 0, S; L3 Td ∩ L2 0, S; H 2 Td ,

2



− 2 ρλ − ηλ dx = C(x) dx, Vλ ∈ L3 0, S; H 2 Td , Td

Td



0, S; H −2 Td , 9 −

1,min{( 32 )− ,( β+4 ) } 0, S; H −2 Td , ηλ2 ∈ W ρλ2 ∈ W

9 − 1,min{( 32 )− ,( α+4 ) }

+ ,( 18 )+ } 9−2α

for any ϕ ∈ Lmax{3

S



∂s ρλ2 , ϕ



i,j =1 0



∂s ηλ2 , ψ



18 + max{3+ ,( 9−2β ) }

(0, S; H 2 (Td )),

S d S

 2 2α  λ2 2 2 λ ∇ρλ − ρλ2 ∇Vλ ∇ϕ dx ds, ds = − ρλ ln ρλ x x ϕxi xj dx ds − i j 2

0

S

(0, S; H 2 (Td )) and ψ ∈ L

d λ2 ds = − 2

Td

S

0

0 Td

i,j =1 0 d T

(1.4)

ηλ2

2

ln ηλ x x ψxi xj dx ds −

S

i j

 2 2β  λ ∇ηλ + ηλ2 ∇Vλ ∇ψ dx ds,

(1.5)

0 Td

and Vλ = ρλ2 − ηλ2 − C(x) Case ii: d = 3, 1  α, β <

a.e. in QS .

(1.6)

16 5 .





16 − 0  ρλ , ηλ ∈ L∞ 0, S; L( 5 ) Td ∩ L2 0, S; H 2 Td ,

2



32 − ρλ − ηλ2 dx = C(x) dx, Vλ ∈ L( 15 ) 0, S; H 2 Td , Td

Td



0, S; H −2 Td , 64

− − 1,min{( 64 47 ) ,( 15β+16 ) } 0, S; H −2 Td ηλ2 ∈ W 64 − ,(

ρλ2 ∈ W 1,min{( 47 )

64 − 15α+16 ) }

64 + ,(

and (ρλ , ηλ , Vλ ) satisfies (1.4)–(1.6) provided that ϕ ∈ Lmax{( 17 ) 64 + + max{( 64 17 ) ,( 48−15β ) }

L

(0, S; H 2 (Td )).

64 + 48−15α ) }

(0, S; H 2 (Td )) and ψ ∈

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

67

Remark 1.1. For the solution (ρλ , ηλ , Vλ ) in Theorem 1.1, letting

  Vλ = Vλ − Vλ dx/Td , Td

we have that (ρλ , ηλ , Vλ ) is also a weak solution satisfying

 Td

Vλ dx = 0.

Corollary 1.1. Let d = 2, 1  α, β < 92 or d = 3, 1  α, β < 16 5 . For those solutions satisfying in Theorem 1.1, {(ρλ , ηλ , Vλ )}λ>0 , we have the following estimate uniformly in λ  2  ρ − η2  2 + Vλ  2  C, 2 d λ

L (0,S;H (T ))

λ L (QS )

 Td

Vλ dx = 0 obtained (1.7)

and moreover, Case i: d = 2, 1  α, β < 92 .

 4 + λ( 3 ) ρλ2 D 2 ln ρλ2  ( 3 )− + ρλ L∞ (0,S;L3− (Td )) L 2 (0,S;L2 (Td ))   2α+9 + + λ( 9 ) ρλ2α  ( 18 )−  C, 1,( 18 )− d 2α+9

L



 4 + λ( 3 ) η2 D 2 ln η2  λ

+ λ(

λ

2β+9 + 9 )

 2β  η  λ

( 32 )−

L

2α+9

(0,S;W

(0,S;L2 (Td ))

18 )− ( 2β+9

L

(0,S;W

(T ))

+ ηλ L∞ (0,S;L3− (Td ))

18 )− 1,( 2β+9

(Td ))

 C.

(1.9)

Case ii: d = 3, 1  α, β < 16 5 . 

 47 + + ρλ  ∞ λ( 32 ) ρλ2 D 2 ln ρλ2  ( 64 )− ( 16 )− L 47 (0,S;L2 (Td )) L (0,S;L 5 (Td ))   15α+16 + + λmax{1,( 32 ) } ρ 2α   C, 64 6 − min{2,( 15α+16 ) }

λ



 47 +  λ( 32 ) ηλ2 D 2 ln ηλ2  +λ

+ max{1,( 15β+16 32 ) }

L

− ( 64 47 )

L

(0,S;L2 (Td ))

 2α  η  λ

1, 5

(0,S;W

+ ηλ 

64 )− } min{2,( 15β+16

L

(1.8)

(Td ))

− ( 16 5 )

L∞ (0,S;L

(0,S;W

1, 56

(Td ))

(1.10)

(Td ))

 C.

(1.11)

 Theorem 1.2. Let d = 2, 1  α, β < 92 or d = 3, 1  α, β < 16 5 . For those solutions satisfying Td Vλ dx = 0 obtained in Theorem 1.1, {(ρλ , ηλ , Vλ )}λ>0 , as λ → 0, there exists a subsequence which is not relabeled, such that ρλ2 − ηλ2  n − p in L2 (QS ),

Vλ  V in L2 0, S; H 2 Td ,

(1.12) (1.13)

and moreover, Case i: d = 2, 1  α, β < 92 .



3 − in L∞ 0, S; L( 2 ) Td ,

− ∇Vλ → ∇V in L2 0, S; L∞ Td ,

3 − 18 − ∂s ρλ2  ∂s n in Lmin{( 2 ) ,( 2α+9 ) } 0, S; H −2 Td , 18 −

min{( 32 )− ,( 2β+9 ) } ∂s ηλ2  ∂s p in L 0, S; H −2 Td , ∗

ρλ2  n,

∗

ηλ2  p

λ2 ∇ρλ2α → 0 in L

18 − ( 2α+9 )

(1.15) (1.16) (1.17) (1.18)

(QS ),

( 18 )− λ ∇ηλ2α → 0 in L 2β+9 (QS ),



λ2 ρλ2 D 2 ln ρλ2 , λ2 ηλ2 D 2 ln ηλ2 → 0

(1.14)

2

in L

( 32 )−





0, S; L2 Td ,

(1.19) (1.20)

68

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

and for any ϕ, ψ ∈ L∞ (0, S; H 2 (Td )), ⎧

S

S ⎪ ⎪ ⎪ ⎪ ⎪ ∂s n, ϕ ds = n∇V ∇ϕ dx ds, ⎪ ⎪ ⎪ ⎪ ⎪ 0 Td ⎨0 S

S ⎪ ⎪ ⎪ ∂s p, ψ ds = − p∇V ∇ψ dx ds, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 0 Td ⎪ ⎩ V = n − p − C(x) a.e. in QS ,

(1.21)

where n, p  0. Case ii: d = 3, 1  α, β <

16 5 .



8 − in L∞ 0, S; L( 5 ) Td ,

− ∇Vλ → ∇V in L2 0, S; L6 Td ,

64 − 64 − ∂s ρλ2  ∂s n in Lmin{( 47 ) ,( 15α+16 ) } 0, S; H −2 Td , 64

− − min{( 64 47 ) ,( 15β+16 ) } 0, S; H −2 Td ∂s ηλ2  ∂s p in L ,

64 6 − λ2 ∇ρλ2α → 0 in Lmin{2,( 15α+16 ) } 0, S; L 5 Td , 64

6 min{2,( 15β+16 )− } λ2 ∇ηλ2α → 0 in L 0, S; L 5 Td ,





64 − λ2 ρλ2 D 2 ln ρλ2 , λ2 ηλ2 D 2 ln ηλ2 → 0 in L( 47 ) 0, S; L2 Td , ∗

ρλ2  n,

∗

ηλ2  p

(1.22) (1.23) (1.24) (1.25) (1.26) (1.27) (1.28)

and (n, p, V ) satisfies (1.21). Remark 1.2. In Theorem 1.2, (n, p, V ) is a weak solution to the initial periodic boundary problem of ⎧ ⎨ ns = − div(n∇V ), ps = div(p∇V ), ⎩ V = n − p − C(x)

(1.29)

in the sense of (1.21). So it shows that the weak solution of QDDM (1.3) converges to that of (1.29) as λ → 0. This article is organized as follows. In Section 2, we show the semi-discretization approximation problem and its solutions. Section 3 contains the uniform entropy estimate which is used in the proof of existence. Then in Section 4, we use a compactness argument for fixed λ > 0 to prove Theorem 1.1. Furthermore in Section 5, we will obtain the limit of λ → 0 by the uniform estimates in λ. 2. Semi-discretization approximate problem The main purpose of this section is to show the approximate problem of (1.3) and its solutions obtained in [25]. More precisely, let τ > 0 such that S = N τ (without loss of generality, otherwise, let N = [ Sτ ] + 1). Hence N N = N(τ ) ∈ N depends only on τ . We divide the time interval (0, S] by (0, S]  = k=1 ((k − 1)τ, kτ ]. For any k = 1, 2, . . . , N , given ρλ,k−1 and ηλ,k−1 such that Td (ρλ,k−1 − ηλ,k−1 ) dx = Td C(x) dx, we will solve the following problem in Td

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

⎧ 2 d 2 ρ − ρλ,k−1



2 ⎪ λ2 2 ⎪ 2 2α ⎪ λ,k = − ρλ,k ln ρλ,k + λ2 ρλ,k − div ρλ,k ∇Vλ,k , ⎪ x x x x ⎪ i j i j τ 2 ⎪ ⎪ i,j =1 ⎪ ⎪ ⎪ ⎪ d ⎪ 2 2 2 ⎪ 2 2

2 ⎪ 2β ⎨ ηλ,k − ηλ,k−1 = − λ ηλ,k ln ηλ,k x x x x + λ2 ηλ,k + div ηλ,k ∇Vλ,k , i j i j τ 2 i,j =1 ⎪ ⎪ ⎪ ⎪ 2 2 ⎪ Vλ,k = ρλ,k − ηλ,k − C(x), ⎪ ⎪ ⎪

⎪ ⎪ 2

⎪ 2 ⎪ ρ dx = − η C(x) dx. ⎪ λ,k λ,k ⎪ ⎩ Td

69

(2.1)

Td

Theorem 2.1. Let d = 2, 3, 1  α, β < ∞ and C(x) ∈ L∞ (Td ). Suppose that ρλ,k−1 and ηλ,k−1 are nonnegative measurable functions on Td with

2

2 ρλ,k−1 − ηλ,k−1 dx = C(x) dx Td

and

Td

 2

 2 ρλ,k−1 ln ρλ,k−1 − 1 + 1 dx < ∞,

Td

 2 2

 ηλ,k−1 ln ηλ,k−1 − 1 + 1 dx < ∞.

Td

2 (Td ))3 satisfying Then there exists a constant cλ,k > 0 such that   (2.1) has a weak solution (ρλ,k , ηλ,k , Vλ,k ) ∈ (H 2 2 2 ρλ,k , ηλ,k  cλ,k > 0, Td (ρλ,k − ηλ,k ) dx = Td C(x) dx and in the sense that, for any ϕ, ψ ∈ H (Td ),

2 − ρ2 ρλ,k λ,k−1

τ

ϕ dx = −

d

2 2α

λ2 2 2 2 ln ρλ,k λ ∇ρλ,k − ρλ,k ρλ,k ϕ dx − ∇Vλ,k ∇ϕ dx, x x i j x x i j 2 i,j =1

Td

2 ηλ,k

2 − ηλ,k−1

τ

ψ dx = −

Td

d λ2

2

Td

i,j =1

Td

2

2 ln ηλ,k x x ψxi xj dx − ηλ,k

i j

Td

(2.2)

2 2β

2 λ ∇ηλ,k + ηλ,k ∇Vλ,k ∇ψ dx,

(2.3)

Td

and Vλ,k is a strong solution of 2 2 Vλ,k = ρλ,k − ηλ,k − C(x)

in Td .

(2.4)

Proof. Although this theorem has been shown in [25], for convenience of the readers, we show the scratch of the proof. Without loss of generality, let λ = 1, ρk = ρλ,k and ηk = ηλ,k . First we solve the following regularized problem ⎧ d 2 ⎪ ρk2 − ρk−1





 1 2 ⎪ ⎪ ρk ln ρk2 x x x x − δ 2 ln ρk2 + ln ρk2 + ρk2α − div ρk2 ∇Vk , = − ⎪ ⎪ i j i j ⎪ τ 2 ⎪ i,j =1 ⎪ ⎪ ⎪ ⎪ ⎪ d 2 2 ⎪  2 2



⎪ 2β ⎨ ηk − ηk−1 = − 1 ηk ln ηk x x x x − δ 2 ln ηk2 + ln ηk2 + ηk + div ηk2 ∇Vk , i j i j τ 2 i,j =1 ⎪ ⎪ ⎪ ⎪ V = ρ 2 − η2 − C(x), ⎪ ⎪ k k k ⎪ ⎪

⎪

⎪ ⎪ 2 2 ⎪ ρ dx = − η C(x) dx ⎪ k k ⎪ ⎩ Td

Td

in Td by a general version of Leray–Schauder fixed-point theorem (see Theorem B.5, p. 262 in [34]).

(2.5)

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X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

Let (ρ, η) ∈ (W 1,4 (Td ))2 → (L∞ (Td ))2 and σ ∈ [0, 1]. Then there is a unique (up to an additive constant) solution V ∈ H 2 (Td ) for the Poisson problem ⎧ V = ρ 2 − η2 − C(x), ⎪ ⎨

2

2 dx = ρ − η C(x) dx ⎪ ⎩ Td

Td

by standard elliptic theory. We introduce the bilinear forms 

 d 1 2 2α ρ wxi xj ϕxi xj + δ(wϕ + wϕ) + αρ ∇w∇ϕ dx, a1 (w, ϕ) = 2 Td



a2 (v, ψ) =



for any w, ϕ ∈ H 2 Td ,

i,j =1

 d 1 2 η vxi xj ψxi xj + δ(vψ + vψ) + βη2β ∇v∇ψ dx, 2



for any v, ψ ∈ H 2 Td ,

i,j =1

Td

and the linear functionals 

 2 ρ 2 − ρk−1 −ρ 2 ∇V ∇ϕ + ϕ dx, F1 (ϕ) = −σ τ Td



F2 (ψ) = −σ

η ∇V ∇ψ + 2

2 η2 − ηk−1

τ

 ψ dx,



for any ϕ ∈ H 2 Td ,

for any ψ ∈ H 2 Td .

Td

Then for any w ∈ H 2 (Td ), 

 d 1 ρ 2 wx2i xj + δ(w)2 + δw 2 + αρ 2α (∇w)2 dx a1 (w, w) = 2 Td

δ

i,j =1



 (w)2 + w 2 dx  Cδw2H 2 (Td )

Td

and for any w, ϕ ∈ H 2 (Td ), by Hölder inequality, we have   a1 (w, ϕ)  Cw 2 d ϕ 2 d , H (T ) H (T )   F1 (ϕ)  Cϕ 2 d . H (T )

By Lax–Milgram theorem, there exists a unique w ∈ H 2 (Td ) such that a1 (w, ϕ) = F1 (ϕ) for any ϕ ∈ H 2 (Td ). In the same vain, we can also find a unique v ∈ H 2 (Td ) which solves a2 (v, ψ) = F2 (ψ) for any ψ ∈ H 2 (Td ). Thus we w v can define the mapping G : (W 1,4 (Td ))2 × [0, 1] → (W 1,4 (Td ))2 by G((ρ, η), σ ) = (ρ, η) = (e 2 , e 2 ). Obviously, we have (ρ, η) ∈ (H 2 (Td ))2 . Now we investigate the assumptions of Leray–Schauder fixed-point theorem. It is easy to check that G((ρ, η), 0) ≡ (1, 1) for any (ρ, η) ∈ (W 1,4 (Td ))2 . Moreover, it can be easily seen that G is continuous and also compact because of the embedding (H 2 (Td ))2 → → (W 1,4 (Td ))2 . To show the uniform bound of fixed-points, let ((ρ, η), σ ) ∈ (W 1,4 (Td ))2 × [0, 1] such that G((ρ, η), σ ) = (ρ, η) ∈ (H 2 (Td ))2 , i.e. (ρ, η, V ) ∈ (H 2 (Td ))3 , where for the sake of explicitness, we relabel it as (ρδ , ηδ , Vδ ), is the solution of ⎧ ⎪ V = ρδ2 − ηδ2 − C(x), ⎪ ⎨ δ

2

2 ρ dx = − η C(x) dx δ δ ⎪ ⎪ ⎩ Td

Td

and for any ϕ, ψ ∈ H 2 (Td ),

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

σ

2 ρδ2 − ρk−1

τ

ϕ dx = − Td

−δ

σ

Td

ηδ2

2 − ηk−1

τ

 d

1 2 2 2α 2 ρδ ln ρδ x x ϕxi xj + ∇ρδ ∇ϕ − σρδ ∇Vδ ∇ϕ dx i j 2



Td

i,j =1



  ln ρδ2 ϕ + ϕ ln ρδ2 dx,

 ψ dx = −

Td

Td

−δ

71

(2.6)

 d 1 2 2

2β 2 ηδ ln ηδ x x ψxi xj + ∇ηδ ∇ψ + σ ηδ ∇Vδ ∇ψ dx i j 2 i,j =1

 2

  ln ηδ ψ + ψ ln ηδ2 dx.

(2.7)

Td

Using ln ρδ2 and ln ηδ2 ∈ H 2 (Td ) as test functions for (2.6) and (2.7) respectively, we deduce, with similar arguments as in Lemma 3.1 and Theorem 3.1 in the following Section 3, that

2

2



 σ  2 σ  2 2 ρδ ln ρδ2 − 1 + 1 dx + ηδ ln ηδ − 1 + 1 dx + σ ρδ − ηδ2 dx τ τ Td

1 + 2

Td

+δ

Td

 



  ρδ D 2 ln ρ 2 2 + ηδ D 2 ln η2 2 dx + 4 δ δ α    

  ln ρ 2 2 + ln ρ 2 2 dx + δ δ

Td

σ  τ

δ

 2

 σ 2 ρk−1 ln ρk−1 − 1 + 1 dx + τ

Td



 α 2 ∇ρ  dx + 4 δ β

Td

 2 2  2 2   ln η  + ln η  dx δ

Td

Td

 β 2 ∇η  dx δ

Td

δ

 2 2

 ηk−1 ln ηk−1 − 1 + 1 dx + C

Td

2

ρδ + ηδ2 dx

(2.8)

Td

and hence the uniform estimates in δ,  



 1 ρδ ; ηδ H 2 (Td ) + ρδ D 2 ln ρδ2 ; ηδ D 2 ln ηδ2 L2 (Td ) + δ 2 ln ρδ2 ; ln ηδ2 H 2 (Td ) + Vδ H 2 (Td )  C

(2.9)

where C is a constant independent of δ. This establishes the uniform bound of fixed-points. So we obtain a solution (ρδ , ηδ ) to G((ρδ , ηδ ), 1) = (ρδ , ηδ ), hence the weak solution of (2.5), by Leray–Schauder fixed-point theorem for any δ > 0. Moreover, we also have the same uniform estimates as (2.9) by repeating the proofs above provided taking σ = 1 in (2.8). By noting the compact embedding H 2 (Td ) → → W 1,4 (Td ), we establish the convergent results as follows, for a subsequence which is not relabeled:

ρδ  ρ; ηδ  η; Vδ  V in H 2 Td ,

ρδ → ρ; ηδ → η; Vδ → V in W 1,4 Td ,





ρδ D 2 ln ρδ2  ρD 2 ln ρ 2 in L2 Td ,





ηδ D 2 ln ηδ2  ηD 2 ln η2 in L2 Td ,

δ ln ρδ2 → 0; δ ln ηδ2 → 0 in H 2 Td . Passing to the limit δ → 0 in the weak form of (2.5), we deduce that ⎧ V = ρ 2 − η2 − C(x), ⎪ ⎨

2

2 ρ dx = − η C(x) dx ⎪ ⎩ Td

Td

and for any ϕ, ψ ∈ H 2 (Td ),

72

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

2 ρ 2 − ρk−1

τ

 ϕ dx = −

Td

Td 2 η2 − ηk−1

τ



ψ dx = −

Td

Td

This finishes the proof.

 d

1 2 2 2α 2 ρ ln ρ x x ϕxi xj + ∇ρ ∇ϕ − ρ ∇V ∇ϕ dx, i j 2 i,j =1

 d 1 2 2

2β 2 η ln η x x ψxi xj + ∇η ∇ψ + η ∇V ∇ψ dx. i j 2 i,j =1

2

Remark 2.1. In the proof of Theorem 2.1, the authors employ the idea of [17] where the approximate problem of a fourth order equation was treated and the exponential transformation technique as in [13–15,22] was also used to ensure the positivity of approximate solution which is essential for the uniform estimate. Remark 2.2. For the solution (ρλ,k , ηλ,k , Vλ,k ) in Theorem 2.1, letting

  Vλ,k = Vλ,k − Vλ,k dx/Td , Td

we have that (ρλ,k , ηλ,k , Vλ,k ) is also a weak solution satisfying

 Td

Vλ,k dx = 0.

3. Uniform estimates of approximate solution Suppose ρ0 , η0 and C(x) satisfy the assumption of Theorem 2.1 for k = 1. We use Theorem 2.1 iteratively to obtain a sequence of approximate solutions (ρλ,k , ηλ,k , Vλ,k ) ∈ (H 2 (Td ))3 (k = 1, 2, . . . , N ). In this section, we focus on the uniform estimates for the approximate solution. From now on, C (or Cλ ) is supposed to be a constant dependent only on S, C(·)L∞ (Td ) , ρ02 (ln ρ02 − 1) + 1L1 (Td ) , η02 (ln η02 − 1) + 1L1 (Td ) (and λ). Lemma 3.1. (Chen [25].) For 1  α, β < ∞, we have

 2  2 2



 2 ρλ,k ln ρλ,k ηλ,k ln ηλ,k − 1 + 1 dx − 1 + 1 dx + Td

+ +



λ2 τ 2

Td

 



  ρλ,k D 2 ln ρ 2 2 + ηλ,k D 2 ln η2 2 dx λ,k λ,k

Td

2 4λ τ

α

2  α 2 ∇ρ  dx + 4λ τ λ,k β

Td

Td

 β 2 ∇ρ  dx + τ λ,k

  2 2 − 1 + 1 dx + ρλ,k−1 ln ρλ,k−1

Td



2

2 2 ρλ,k − ηλ,k dx

Td

  2 2 − 1 + 1 dx + Cτ ηλ,k−1 ln ηλ,k−1

Td



2 2 + ηλ,k ρλ,k dx.

(3.1)

Td

2 , ln η2 ∈ Proof. Using the Sobolev embedding H 1 (Td ) → L4 (Td ) and ρλ,k , ηλ,k  cλ,k > 0, we have ln ρλ,k λ,k H 2 (Td ), and hence finish the proof by employing them as test functions in (2.2) and (2.3) respectively. 2

Lemma 3.2. For γ > 1 and 1  α, β < ∞, we have

2γ

τ 2γ 1 2γ

2γ 2 2 ρλ,k + ηλ,k dx + ρλ,k − ηλ,k ρλ,k dx − ηλ,k γ (γ − 1) γ Td

+

Td

d  2

2(γ −1)

2 2(γ −1)  λ2 τ 2 2 ρλ,k ln ρλ,k ρλ,k ln ηλ,k x x ηλ,k dx + ηλ,k x x x x xi xj i j i j i j 2(γ − 1) i,j =1

Td

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

4λ2 ατ (α + γ − 1)2

× 

1 γ (γ − 1)



 α+γ −1 2 ∇ρ  dx + λ,k

Td

4λ2 βτ (β + γ − 1)2

2γ

2γ ρλ,k−1 + ηλ,k−1 dx + Cτ

Td



73

 β+γ −1 2 ∇η  dx λ,k

Td

2γ 2γ

ρλ,k + ηλ,k dx.

(3.2)

Td 2(γ −1)

Proof. We could establish (3.2) by employing tively. 2

ρλ,k γ −1

2(γ −1)

,

ηλ,k γ −1

∈ H 2 (Td ) as test functions in (2.2) and (2.3) respec-

Remark 3.1. In Lemma 3.2, the test functions were once used in the study of fourth order parabolic equation and drift-diffusion model such as in [17,33] etc. While for QDDM, they are used for the first time up to the authors’ knowledge. Definition 3.1. Define the piecewise constant function in s in the following sense:  ρλ,τ (s, x)  ρλ,k (x) for x ∈ Td , s ∈ (k − 1)τ, kτ ,  ηλ,τ (s, x)  ηλ,k (x) for x ∈ Td , s ∈ (k − 1)τ, kτ ,  Vλ,τ (s, x)  Vλ,k (x) for x ∈ Td , s ∈ (k − 1)τ, kτ , where (ρλ,k , ηλ,k , Vλ,k ) ∈ (H 2 (Td ))3 is the solution in Theorem 2.1, k = 1, 2, . . . , N . Remark 3.2. In Definition 3.1, 0 < ρλ,τ (s, ·), ηλ,τ (s, ·) ∈ H 2 (Td ) for any s ∈ (0, T ]. Using Lemmas 3.1 and 3.2, we now obtain the uniform estimates in τ for ρλ,τ , ηλ,τ , Vλ,τ as follows. Theorem 3.1. For 1  γ <

(d = 2, 3) and 1  α, β < ∞, we have  2  2  ρλ,τ L∞ (0,S;L2γ (Td )) + ηλ,τ L∞ (0,S;L2γ (Td )) + ρλ,τ − ηλ,τ L2 (QS )  α+γ −1   β+γ −1   2  2 + λ∇ρλ,τ + λ∇ηλ,τ L (QS ) L (QS )  

 2  2 2   2   + λ ρλ,τ D ln ρλ,τ L2 (Q ) + λ ηλ,τ D 2 ln ηλ,τ L (QS ) S  γ   γ      + λ ρλ,τ L2 (0,S;H 2 (Td )) + λ ηλ,τ L2 (0,S;H 2 (Td ))  C. 2d+2 d+2

(3.3)

Moreover, we have Case i: d = 2, 1  α, β < 3γ . 1

1

λ 2γ ρλ,τ L4γ (0,S;L∞ (Td )) + λ 2γ ηλ,τ L4γ (0,S;L∞ (Td )) +λ

2−γ 4γ

ρλ,τ 

8γ L 2−γ

(0,S;L4 (Td ))

1

+λ

2−γ 4γ

ηλ,τ 

1

+ λ 3γ ρλ,τ L6γ (QS ) + λ 3γ ηλ,τ L6γ (QS ) 1+2γ 

 2 2  + λ 2γ ρλ,τ D 2 ln ρλ,τ 4γ +λ +λ

1+2γ +α 3γ 3γ +α 3γ

1 γ

8γ

L 2−γ (0,S;L4 (Td ))

 2α  ∇ρ  λ,τ

 2α  ρ  λ,τ

L 1+2γ

6γ L 1+2γ +α

6γ L 3γ +α

(0,S;W

(QS )

(0,S;L2 (Td ))

+λ

6γ 1, 3γ +α

+ λ Vλ,τ L2γ (0,S;H 2 (Td ))  C.

1+2γ +β 3γ

(Td ))

+λ

+λ

1+2γ 2γ

 2

 η D 2 ln η2  λ,τ λ,τ

 2β  ∇η  λ,τ

3γ +β 3γ

4γ

L 1+2γ (0,S;L2 (Td ))

6γ

L 1+2γ +β (QS )

 2β  η  λ,τ

6γ

L 3γ +β (0,S;W

6γ 1, 3γ +β

(Td ))

(3.4)

74

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

Case ii: d = 3, 1  α, β < 2γ . 3

λ 4γ ρλ,τ  6−3γ 8γ

+λ

3

8γ L 3

(0,S;L∞ (Td ))

ρλ,τ 

16γ

L 6−3γ (0,S;L4 (Td ))

3

+ λ 7γ ρλ,τ  3+4γ 4γ

+λ +λ +λ

+ λ 4γ ηλ,τ  +λ

L

8γ 3

6−3γ 8γ

(0,S;L∞ (Td ))

ηλ,τ 

16γ

L 6−3γ (0,S;L4 (Td ))

3

14γ L 3

(QS )

+ λ 7γ ηλ,τ 

 2

 ρ D 2 ln ρ 2  λ,τ λ,τ

8γ L 3+4γ

L

14γ 3

(QS )

(0,S;L2 (Td ))

+λ

3+4γ 4γ

 2

 η D 2 ln η2  λ,τ λ,τ

+3α  max{1, 2γ4γ }  2α  ρλ,τ  min{2, 8γ } 6 2γ +3α (0,S;W 1, 5 (Td )) L

8γ

L 3+4γ (0,S;L2 (Td ))

+3β  max{1, 2γ4γ }  2β  ηλ,τ  min{2, 8γ } 6 2γ +3β (0,S;W 1, 5 (Td )) L 3

+ λ 2γ Vλ,τ 

L

4γ 3

(0,S;H 2 (Td ))

 C.

(3.5)

Proof. First, we establish (3.3) with similar discussion as in [25]. By Lemmas 3.1–3.2, the Gronwall inequality and the inequality x  x(ln x − 1) + 3,

for any x > 0,

we have ρλ,τ L∞ (0,S;L2γ (Td )) + ηλ,τ L∞ (0,S;L2γ (Td ))  C, and hence  α+γ −1   β+γ −1    2   2  2 ρ − η2  2 + λ∇ηλ,τ λ,τ λ,τ L (QS ) + λ ∇ρλ,τ L (QS ) L (QS )  

 2  2 2  2    + λ ρλ,τ D ln ρλ,τ L2 (Q ) + λ ηλ,τ D ln ηλ,τ L2 (Q ) S S  d   d     







    2(γ −1) 2(γ −1) 2 2 2 2 2 + λ2  ρλ,τ + λ η ln ρλ,τ ρ ln η η   λ,τ λ,τ λ,τ λ,τ xi xj xi xj  xi xj xi xj     1 i,j =1

i,j =1

L (QS )

Using Lemma 4 in [17], one has  γ   γ  + λη  λ ρ  2 2 d

λ,τ L2 (0,S;H 2 (Td ))

λ,τ L (0,S;H (T ))

L1 (QS )

 C.

So (3.3) is proved for d = 2, 3. In the following, we prove (3.4) and (3.5) by using delicate interpolation skill. Case i: By the Gagliardo–Nirenberg inequality, we have  1  1  γ  ρ  ∞ d  C ρ γ  2 2 d ρ γ  2 2 d , λ,τ L (T ) λ,τ H (T ) λ,τ L (T )  γ  ρ  λ,τ

4 Lγ

 γ  ρ  λ,τ

(Td )

L6 (Td )

 γ  2−γ  γ  2+γ  C ρλ,τ H42 (Td ) ρλ,τ L24(Td ) ,

 γ 1  γ 2  C ρλ,τ H3 2 (Td ) ρλ,τ L3 2 (Td ) .

These formulas and Hölder inequality, together with (3.3) imply  γ  12 1 γ  1  γ 1 ρ  ∞ λ 2 ρ  4  Cλ 2 ρ  2 2 ∞ d 2 d λ,τ L (0,S;L (T ))

λ

2−γ 4

 γ  ρ  λ,τ

8 L 2−γ

4 (0,S;L γ

λ,τ L (0,S;H (T ))

(Td ))

 Cλ

2−γ 4

 γ  2−γ ρ  24

λ,τ L (0,S;L2 (Td ))

λ,τ L (0,S;H 2 (Td ))

 γ  2+γ 4 ρ  ∞

λ,τ L (0,S;L2 (Td ))

 γ 2 1 γ  1  γ 1 λ 3 ρλ,τ L6 (Q )  Cλ 3 ρλ,τ L3 2 (0,S;H 2 (Td )) ρλ,τ L3 ∞ (0,S;L2 (Td ))  C. S

 C,  C,

 C.

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

75

Hence 1

λ 2γ ρλ,τ L4γ (0,S;L∞ (Td )) + λ

2−γ 4γ

ρλ,τ 

1

8γ L 2−γ

(0,S;L4 (Td ))

+ λ 3γ ρλ,τ L6γ (QS )  C

and by (3.3) that λ

1+2γ 2γ

 2

 ρ D 2 ln ρ 2  λ,τ λ,τ

4γ L 1+2γ

(0,S;L2 (Td ))

 1

 2   λ 2γ ρλ,τ L4γ (0,S;L∞ (Td )) λρλ,τ D 2 ln ρλ,τ  C. L2 (Q ) S

From chain rule and Hölder inequality, one has λ

λ

1+2γ +α 3γ

3γ +α 3γ

 2α  ∇ρ  λ,τ

6γ L 1+2γ +α

 2α  ∇ρ  λ,τ

6γ

λ,τ

L 3γ +α (QS )

With the help of  2α  λ 3γ ρ 2α 

L

(QS )

α

λ,τ L (QS )

L (QS )

λ,τ L2 (QS )

 C.

2α

6γ L 2α

λ,τ

1+2γ +α   α+γ −1  2α α−γ +1   6γ ∇ρ  2 λ 3γ ρλ,τ λ,τ L (QS ) α−γ +1 α + γ − 1 (QS ) L (QS ) α−γ +1 2α α−γ +1  α+γ −1   2 λ 3γ ρλ,τ L6γ (Q ) λ∇ρλ,τ =  C, L (QS ) S α+γ −1 3γ +α     α   α ∇ρ  2  2λ 3γ ρ α  6γ = 2λ 3γ ρλ,τ α 6γ λ∇ρ α 



we deduce that 3γ +α   2α  λ 3γ ρλ,τ

(QS )

= λ 3γ ρλ,τ 2α  C, L6γ (Q ) S

6γ

L 3γ +α (0,S;W

6γ 1, 3γ +α

(Td ))

 C.

In the same vain, we could establish the corresponding estimates about ηλ,τ and therefore the estimate of Vλ,τ by standard elliptic estimates. Case ii: With similar argument as in Case i, one could deduce (3.5). For explicitness, we show a part of the details in the following. From the Gagliardo–Nirenberg inequality, it holds that  γ  ρ 

 γ 3  γ 1  C ρλ,τ H4 2 (Td ) ρλ,τ L4 2 (Td ) ,

 γ  ρ 

 γ  6−3γ  γ  2+3γ  C ρλ,τ H 28 (Td ) ρλ,τ L28(Td ) ,

λ,τ L∞ (Td ) λ,τ

4 Lγ

(Td )

 γ  ρ  λ,τ

14 L 3 (Td )

 γ 3  γ 4  C ρλ,τ H7 2 (Td ) ρλ,τ L7 2 (Td ) .

So we have, in view of Hölder inequality and (3.3) that 3 γ  λ 4 ρλ,τ 

8 L 3 (0,S;L∞ (Td ))

λ

6−3γ 8

 γ  ρ  λ,τ

16 L 6−3γ

3 γ  λ 7 ρλ,τ 

14 L 3 (QS )

 γ 1 3  γ 3  Cλ 4 ρλ,τ L4 2 (0,S;H 2 (Td )) ρλ,τ L4 ∞ (0,S;L2 (Td ))  C,

4 (0,S;L γ

(Td ))

 Cλ

6−3γ 8

 γ  6−3γ ρ  28

λ,τ L (0,S;H 2 (Td ))

 γ  2+3γ ρ  ∞8

λ,τ L (0,S;L2 (Td ))

 C,

 γ 4 3  γ 3  Cλ 7 ρλ,τ L7 2 (0,S;H 2 (Td )) ρλ,τ L7 ∞ (0,S;L2 (Td ))  C.

Thus 3

λ 4γ ρλ,τ 

8γ L 3

(0,S;L∞ (Td ))

+λ

6−3γ 8γ

ρλ,τ 

In the following, we establish the estimate +3α  max{1, 2γ4γ }  2α  ρλ,τ  min{2, 8γ } λ 1, 6 L

2γ +3α

(0,S;W

3

16γ L 6−3γ

5 (Td ))

(0,S;L4 (Td ))

 C.

+ λ 7γ ρλ,τ 

L

14γ 3

(QS )

 C.

(3.6)

76

If

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80 2γ 3

< α < 2γ , by Gagliardo–Nirenberg inequality again, one could deduce that λ

3α−2γ 4α

 γ  ρ 

8α L 3α−2γ

λ,τ

3α (0,S;L γ

(Td ))

 Cλ

3α−2γ 4α

 γ  3α−2γ ρ  24α

3α−2γ 4γ

 α  ρ 

λ,τ L (0,S;H 2 (Td ))

 γ  α+2γ 4α ρ  ∞

λ,τ L (0,S;L2 (Td ))

C

and hence λ

3α−2γ 4αγ

ρλ,τ 

8αγ

L 3α−2γ (0,S;L3α (Td ))

 C.

Therefore λ

2γ +3α 4γ

 2α  ∇ρ  λ,τ

8γ L 2γ +3α

6 (0,S;L 5 (Td ))

 2λ = 2λ

3α−2γ 4γ

λ,τ

8γ L 3α−2γ

(0,S;L3 (Td ))

 α   2 λ∇ρλ,τ L (Q

ρλ,τ α

8αγ L 3α−2γ

(0,S;L3α (Td ))

S)

 α   2 λ∇ρλ,τ C L (Q ) S

and λ

3α−2γ 2γ

 2α  ρ  λ,τ

8γ L 2γ +3α

6 (0,S;L 5 (Td ))

=λ

3α−2γ 2γ

 Cλ

ρλ,τ 2α 16αγ

L 2γ +3α (0,S;L

3α−2γ 2γ

12α 5

(Td ))

ρλ,τ 2α 8αγ

 C.

L 3α−2γ (0,S;L3α (Td ))

Combing the above two formulas, we obtain (3.6) for 2γ 3 < α < 2γ . While for 1  α   2α   α  α ∇ρ  2 λ∇ρλ,τ  2  2λρ  6 λ,τ ∞ 3α d λ,τ L (QS ) L (0,S;L (T )) L (0,S;L 5 (Td ))   α   Cλρλ,τ αL∞ (0,S;L2γ (Td )) ∇ρλ,τ C L2 (Q )

2γ 3

, it is self-evident that

S

and

 2α   2 = λρλ,τ 2α  Cλρλ,τ 2α8γ λρλ,τ L (Q ) L4α (Q ) S

S

So in this case, (3.6) is also valid.

L

3

 C. (QS )

2

4. Existence of weak solution Throughout this section, let ρλ,τ , ηλ,τ , Vλ,τ be the functions in Definition 3.1, which satisfy Theorem 3.1. Using a compactness argument and Aubin–Lions lemma (see [27]), we can prove the following convergent results in Theorem 4.1 which will complete the proof of Theorem 1.1. 2 by Definition 4.1. We define the difference quotient of ρλ,τ  2 2 ρλ,k (x)−ρλ,k−1 (x) for x ∈ Ω, s ∈ ((k − 1)τ, kτ ], τ 2 τ ∂s ρλ,τ (s, x)  2 for x ∈ Ω, s = 0, ρ0 (x) 2 . and similarly the difference quotient of ηλ,τ

Theorem 4.1. For any fixed λ > 0 and 1  α, β < ∞, as τ → 0, there exists a subsequence of {(ρλ,τ , ηλ,τ , Vλ,τ , 2 , ∂ τ η2 )} ∂sτ ρλ,τ s λ,τ τ >0 which is not relabeled, such that ρλ,τ  ρλ , and moreover,

ηλ,τ  ηλ



in L2 0, S; H 2 Td ,

(4.1)

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

Case i: d = 2, 1  α, β < 92 .

− Vλ,τ  Vλ in L3 0, S; H 2 Td ,

∗ ∗ − ρλ,τ  ρλ , ηλ,τ  ηλ in L∞ 0, S; L3 Td ,

10 − Vλ,τ → Vλ in L( 7 ) 0, S; H 2 Td ,

10 − − 2 2 ρλ,τ (0 < θ < 1), → ρλ2 , ηλ,τ → ηλ2 in L( 7 ) 0, S; C 0,θ ∩ W 1,∞ Td 2 ρλ,τ

→ ρλ2 ,

→ ηλ2

2 ηλ,τ

18 − ( 2α+9 )

in L

( 92 )−

(QS ),



18 − 1,( 2α+9 ) d

0, S; W T , 18

− 2β 2β 1,( ) ηλ,τ  ηλ in L 0, S; W 2β+9 Td ,

3 − 9 − 2  ∂s ρλ2 in Lmin{( 2 ) ,( α+4 ) } 0, S; H −2 Td , ∂sτ ρλ,τ 9 −

min{( 32 )− ,( β+4 ) } 2 0, S; H −2 Td ,  ∂s ηλ2 in L ∂sτ ηλ,τ





3 − 2 2 ρλ,τ  ρλ2 D 2 ln ρλ2 in L( 2 ) 0, S; L2 Td , D 2 ln ρλ,τ 2





3 − 2 ηλ,τ  ηλ2 D 2 ln ηλ2 in L( 2 ) 0, S; L2 Td , D 2 ln ηλ,τ 2α ρλ,τ

 ρλ2α

in L

18 − ( 2β+9 )

77

(4.2) (4.3) (4.4) (4.5) (4.6) (4.7) (4.8) (4.9) (4.10) (4.11) (4.12)

where ρλ , ηλ  0. Case ii: d = 3, 1  α, β <

16 5 .



32 − in L( 15 ) 0, S; H 2 Td ,

16 − ∗ ∗ ρλ,τ  ρλ , ηλ,τ  ηλ in L∞ 0, S; L( 5 ) Td ,

46 − Vλ,τ → Vλ in L( 38 ) 0, S; H 2 Td ,   d

1 2 2 2 2 ( 46 )− 0,θ 1,6− 38 , ρλ,τ → ρλ , ηλ,τ → ηλ in L 0<θ < 0, S; C ∩ W T 2 Vλ,τ  Vλ

2 ρλ,τ → ρλ2 ,

2 ηλ,τ → ηλ2

56 −

in L( 15 ) (QS ),

64 6 − 2α ρλ,τ  ρλ2α in Lmin{2,( 15α+16 ) } 0, S; W 1, 5 Td , 64

6 2β 2β min{2,( 15β+16 )− } 0, S; W 1, 5 Td , ηλ,τ  ηλ in L

64 − 64 − 2 ∂sτ ρλ,τ  ∂s ρλ2 in Lmin{( 47 ) ,( 15α+16 ) } 0, S; H −2 Td , 64

− − min{( 64 2 47 ) ,( 15β+16 ) } 0, S; H −2 Td ,  ∂s ηλ2 in L ∂sτ ηλ,τ





64 − 2 2 ρλ,τ  ρλ2 D 2 ln ρλ2 in L( 47 ) 0, S; L2 Td , D 2 ln ρλ,τ 2





64 − 2 ηλ,τ  ηλ2 D 2 ln ηλ2 in L( 47 ) 0, S; L2 Td , D 2 ln ηλ,τ

(4.13) (4.14) (4.15) (4.16) (4.17) (4.18) (4.19) (4.20) (4.21) (4.22) (4.23)

where ρλ , ηλ  0. Proof. With Theorem 3.1 at hand, some proofs are similar to that of [25], only that (4.5) and (4.16) are worthy of being shown in detail. But for the convenience of readers, we also give the proofs of the other formulas. By Remark 3.2 and (3.3) with γ = 1, we deduce that there exist ρλ , ηλ  0 satisfying (4.1). 9 − ) } and 0 < τ < 1 be fixed. Then for any 0 < h < τ , it is easy to deduce that Case i: Let r = min{( 32 )− , ( α+4   1 πh ρ 2 − ρ 2  r r λ,τ λ,τ L (0,S−h;H −2 (Td ))  Cλ h ,

where (πh f )(s) = f (s + h). If given the estimate  2  ρ  10 −  Cλ , λ,τ

L(

7 )

(0,S;H 2 (Td ))

(4.24)

78

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

2 in (4.5) by the Aubin–Lions lemma and that of η2 similarly. Now we we could establish the convergence of ρλ,τ λ,τ 9 prove (4.24). For any fixed 1  α < 2 , there exists 1 < γ < 32 such that 1  α < 3γ . Employing the Gagliardo– Nirenberg inequality and Theorem 3.1, one has

∇ρλ,τ 

γ +2

4γ +4 L γ +2 (0,S;L4 (Td ))

γ

 Cρλ,τ L2γ2+2 ρ  2γ +2  Cλ (0,S;H 2 (Td )) λ,τ L∞ (0,S;L2γ (Td ))

and hence by chain rule,  2  ρ 

2γ +2 L γ +2 (0,S;L2 (Td ))

λ,τ

 2−γ γ   Cλ ρλ,τ ρλ,τ 

4γ L γ +2 (0,S;L2 (Td ))

  + |∇ρλ,τ |2 

2γ +2 L γ +2 (0,S;L2 (Td ))

 γ  2−γ  Cλ ρλ,τ L4γ (0,S;L∞ (Td )) ρλ,τ L2 (Q ) + ∇ρλ,τ 2 4γ +4 S

L

γ +2



 Cλ .

(0,S;L4 (Td ))

Since it is easy to check that  2  ρ  2γ +2  Cλ , λ,τ γ +2

L

we obtain  2  ρ  λ,τ

L

(0,S;H 1 (Td ))

2γ +2 γ +2 (0,S;H 2 (Td ))

 Cλ .

Provided that γ = ( 32 )− , (4.24) is proved. By standard elliptic estimates, we could establish (4.4). In addition, (4.2) and (4.3) can be obtained directly by (3.4) and (3.3), respectively. It is easy to deduce that  τ 2  ∂ ρ  r s λ,τ L (0,S;H −2 (Td ))  Cλ . The mean value theorem of differentials together with (4.5) shows that 2  ∂s ρλ2 ∂sτ ρλ,τ

in D (QS ),

where D (QS ) represents the set of all distributions on QS . Therefore (4.9) is proved and similarly (4.10). From (3.4) and (4.5), we are easy to prove the other formulas in Case i. Case ii: So long as we establish  2  ρ  46 −  Cλ , λ,τ ( ) 2 d L

38

(4.25)

(0,S;H (T ))

8 − (4.16) could be obtained similar as in Case i. To prove (4.25), for any fixed 1  α < 16 5 , we choose γ = ( 5 ) such that 1  α < 2γ . Using the Gagliardo–Nirenberg inequality and Theorem 3.1 again, we deduce that

∇ρλ,τ 

γ +6

4γ +12 L γ +6 (0,S;L4 (Td ))

γ

 Cρλ,τ L2γ2+6 ρ  2γ +6  Cλ . (0,S;H 2 (Td )) λ,τ L∞ (0,S;L2γ (Td ))

Then it follows from the chain rule that  2  ρ  λ,τ

2γ +6 L γ +6 (0,S;L2 (Td ))

 2−γ γ   Cλ ρλ,τ ρλ,τ  2−γ  Cλ ρλ,τ  8γ L

3

8γ L γ +6 (0,S;L2 (Td ))

(0,S;L∞ (Td ))

  + |∇ρλ,τ |2 

 γ  ρ 

λ,τ L2 (QS )

and hence (4.25). The proofs of the other formulas are similar as in Case i.

2γ +6 L γ +6 (0,S;L2 (Td ))

+ ∇ρλ,τ 2 4γ +12 L

2

Proof of Theorem 1.1. It is obvious to prove Theorem 1.1 by Theorems 2.1–4.1.

2

γ +6



(0,S;L4 (Td ))



 Cλ

X. Chen, L. Chen / J. Math. Anal. Appl. 343 (2008) 64–80

79

5. Limit of λ → 0 Let us turn to discuss the limit of λ → 0 for the weak solution (ρλ , ηλ , Vλ ) obtained in Theorem 1.1. Proof of Corollary 1.1. The proof is based on Theorems 3.1 and 4.1 in view of the weakly lower semi-continuity of norm as well as standard elliptic estimates. 2 Proof of Theorem 1.2. Case i: Corollary 1.1 implies (1.13), (1.14) and  2 ∂s ρ   C, λ min{( 3 )− ,( 18 )− } −2 d L

2

2α+9

(0,S;H

(T ))

so that (1.16) is also valid. In the same vain, (1.17) is true and hence one deduces that   ∂s (Vλ )  C. min{( 3 )− ,( 18 )− } −2 d L

2

2α+9

(0,S;H

(T ))

This together with Vλ L2 (0,S;H 2 (Td ))  C implies (1.15) by Aubin–Lions lemma. The proofs of the other formulas could be obtained directly by Corollary 1.1. Letting λ → 0 in (1.4) and (1.5) with the test functions ϕ, ψ ∈ L∞ (0, T ; H 2 (Td )), as well as (1.6), we complete the proof of Case i with the help of Corollary 1.1. Case ii: The proof is similar as in Case i.

2

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